SQUARE ROOT OF A COMPLEX NUMBER
The solution of the equation $$z^2=a+ib$$ is given in two forms.
In terms of $a,b$,
$$ z= \pm \left( \sqrt{{ \sqrt{a^2+b^2}+a \over 2}} + i {b\over |b|}\sqrt{{ \sqrt{a^2+b^2}-a \over 2}}\right)$$
In terms of $w=a+ib$,
$$ z= \pm \left(\sqrt{{ |w|+\Re(w) \over 2}} + i{\Im(w)\over |\Im(w)|} \sqrt{{ |w|-\Re(w) \over 2}}\right)$$
A crazy person could just check this by squaring it! you should get $w=a+ib$ back.
=========================== WORKING BELOW
Suppose we wish to find the square root of a complex number $w=a+ib$.
Some notation we use later regarding real and imaginary parts of the complex number.
$\Re(w)=a$
$\Im(w)=b$
${x\over |x|}$ gives the sign of $x$, $+1$ or $-1$.
Let $z=x+iy$, and of course keep in mind that we use $i^2=-1$.
$$\begin{array}{ll} z^2 &=a+ib\\ (x+iy)^2 &= a+ib\\x^2-y^2+i2xy &= a+ib \end{array}$$
Therefore equating real and imaginary parts we have two equations:
$x^2-y^2=a$ and $2xy=b$.
$x^2-\displaystyle{b^2\over 4x^2}=a$
$4 x^4-b^2=4 x^2 a$
$4x^4-4a x^2 -b^2 = 0$
$x^2 = {4a \pm \sqrt{16a^2-4\times 4(-b^2)} \over 8}$
$x^2 = {4a \pm \sqrt{16a^2+16b^2} \over 8}$
$x^2 = {a \pm \sqrt{a^2+b^2} \over 2}$
$x^2 = {a + \sqrt{a^2+b^2} \over 2}$
since $x^2\ge 0$ needs to hold.
$\therefore x = \pm \sqrt{{a + \sqrt{a^2+b^2} \over 2}}$
$\therefore y= \pm {b\over 2} \sqrt{{2 \over a+\sqrt{a^2+b^2} }}$
rationalise the denominator and bring the '$b/2$' inside the square root symbol by keeping in mind that $b=sign(b)\times |b|$.
$\therefore y= \pm sign(b) \sqrt{{b^2\over 4}\,{2 \over a+\sqrt{a^2+b^2} } \times { a-\sqrt{a^2+b^2} \over a-\sqrt{a^2+b^2}}}$
Finally in terms of $a$ and $b$ we get
$$ z= \pm \sqrt{{ \sqrt{a^2+b^2}+a \over 2}} \pm i {b\over |b|} \sqrt{{ \sqrt{a^2+b^2}-a \over 2}}$$
or in terms of $w$
$$ z= \pm \sqrt{{ |w|+\Re(w) \over 2}} \pm i {\Im(w)\over |\Im(w)|}\sqrt{{ |w|-\Re(w) \over 2}}$$
The $\pm$ means you either take both as $+$ or both as $-$. :-)
Or with a big bracket but looks more cumbersome ..
So in terms of $a,b$,
$$ z= \pm \left( \sqrt{{ \sqrt{a^2+b^2}+a \over 2}} + i {b\over |b|} \sqrt{{ \sqrt{a^2+b^2}-a \over 2}}\right)$$
Or in terms of $w$,
$$ z= \pm \left(\sqrt{{ |w|+\Re(w) \over 2}} + i {\Im(w)\over |\Im(w)|}\sqrt{{ |w|-\Re(w) \over 2}}\right)$$
A crazy person could just check this by squaring it! you should get $w=a+ib$ back.
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Postscript Note:
Apparently it can also be shown that
$$\sqrt{z}=\sqrt{|z|}\,{ z+|z|\over \left| z + |z| \right| }$$
where $\sqrt{z}$ means the principal square root, which is where the real part is positive.
You can prove this formula by taking $z=x+iy$ and brute force it for example :-), enjoy!!
